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In the xy coordinate plane, line L and line K intersect at

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In the xy coordinate plane, line L and line K intersect at  [#permalink]

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In the xy coordinate plane, line L and line K intersect at the point (4,3). Is the product of their slopes negative?

(1) The product of the x-intersects of lines L and K is positive.
(2) The product of the y-intersects of lines L and K is negative.

Originally posted by LM on 06 May 2010, 12:31.
Last edited by Bunuel on 24 May 2012, 12:49, edited 1 time in total.
Edited the question and added the OA
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In the xy coordinate plane, line L and line K intersect at  [#permalink]

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New post 06 May 2010, 13:24
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In the xy coordinate plane, line L and line K intersect at the point (4,3). Is the product of their slopes negative?

We have two lines: \(y_l=m_1x+b_1\) and \(y_k=m_2x+b_2\). The question: is \(m_1*m_2<0\)?

Lines intersect at the point (4,3) --> \(3=4m_1+b_1\) and \(3=4m_2+b_2\)

(1) The product of the x-intersects of lines L and K is positive. Now, one of the lines can intersect x-axis at 0<x<4 (positive slope) and another also at 0<x<4 (positive slope), so product of slopes also will be positive BUT it's also possible one line to intersect x-axis at 0<x<4 (positive slope) and another at x>4 (negative slope) and in this case product of slopes will be negative. Two different answers, hence not sufficient.

But from this statement we can deduce the following: x-intersect is value of \(x\) for \(y=0\) and equals to \(x=-\frac{b}{m}\) --> so \((-\frac{b_1}{m_1})*(-\frac{b_2}{m_2})>0\) --> \(\frac{b_1b_2}{m_1m_2}>0\).

(2) The product of the y-intersects of lines L and K is negative. Now, one of the lines can intersect y-axis at 0<y<3 (positive slope) and another at y<0 (positive slope), so product of slopes will also be positive BUT it's also possible one line to intersect y-axis at y<0 (positive slope) and another at y>3 (negative slope) and in this case product of slopes will be negative. Two different answers, hence not sufficient.

But from this statement we can deduce the following: y-intercept is value of \(y\) for \(x=0\) and equals to \(y=b\) --> \(b_1*b_2<0\).

(1)+(2) \(\frac{b_1b_2}{m_1m_2}>0\) and \(b_1*b_2<0\). As numerator in \(\frac{b_1b_2}{m_1m_2}>0\) is negative, then denominator \(m_1m_2\) must also be negative. So \(m_1m_2<0\). Sufficient.

Answer: C.

In fact we arrived to the answer C, without using the info about the intersection point of the lines. So this info is not needed to get C.

For more on coordinate geometry check the link in my signature.

P.S. This question can be easily solved by drawing the lines without any calculations.
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Re: Interesting Line Eqn DS problem  [#permalink]

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New post 19 Jul 2010, 08:29
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line L: y1 = m1x1 + b1; x-intercept = -b1/m1; y-intercept = b1
line K: y2 = m2x2 + b2; x-intercept = -b2/m2; y-intercept = b2

(4,3) lies on both.
3 = 4m1 + b1 = 4m2 + b2

Is m1m2 < 0?

1. (-b1/m1)*(-b2/m2) > 0
b1b2/m1m2 > 0
If b1b2 > 0, m1m2 > 0
If b1b2 < 0, m1m2 < 0
NOT SUFFICIENT.

2. b1b2 < 0
NOT SUFFICIENT.

Together, b1b2 < 0 and m1m2 < 0.
SUFFICIENT. Answer is C.
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In the xy-coordinate plane, line l and line k intersect at the p  [#permalink]

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New post 07 Sep 2010, 08:29
Bunuel wrote:
M8 wrote:
In the xy-coordinate plane, line l and line k intersect at the point (4,3). Is the product of their slope negative?

1) The product of the x-intercepts of lines l and k is positive.
2) The product of the y-intercepts of lines l and k is negative.

I cracked this question right. :wink:
But these types of questions are real horror for me.

Guys please elaborate your solution for me. Thank you.

I'll post the OA later.


In the xy coordinate plane, line L and line K intersect at the point (4,3). Is the product of their slopes negative?

We have two lines: \(y_l=m_1x+b_1\) and \(y_k=m_2x+b_2\). The question: is \(m_1*m_2<0\)?

Lines intersect at the point (4,3) --> \(3=4m_1+b_1\) and \(3=4m_2+b_2\)

(1) The product of the x-intersects of lines L and K is positive. Now, one of the lines can intersect x-axis at 0<x<4 (positive slope) and another also at 0<x<4 (positive slope), so product of slopes also will be positive BUT it's also possible one line to intersect x-axis at 0<x<4 (positive slope) and another at x>4 (negative slope) and in this case product of slopes will be negative. Two different answers, hence not sufficient.

But from this statement we can deduce the following: x-intersect is value of \(x\) for \(y=0\) and equals to \(x=-\frac{b}{m}\) --> so \((-\frac{b_1}{m_1})*(-\frac{b_2}{m_2})>0\) --> \(\frac{b_1b_2}{m_1m_2}>0\).

(2) The product of the y-intersects of lines L and K is negative. Now, one of the lines can intersect y-axis at 0<y<3 (positive slope) and another at y<0 (positive slope), so product of slopes will also be positive BUT it's also possible one line to intersect y-axis at y<0 (positive slope) and another at y>3 (negative slope) and in this case product of slopes will be negative. Two different answers, hence not sufficient.

But from this statement we can deduce the following: y-intercept is value of \(y\) for \(x=0\) and equals to \(x=b\) --> \(b_1*b_2<0\).

(1)+(2) \(\frac{b_1b_2}{m_1m_2}>0\) and \(b_1*b_2<0\). As numerator in \(\frac{b_1b_2}{m_1m_2}>0\) is negative, then denominator \(m_1m_2\) must also be negative. So \(m_1m_2<0\). Sufficient.

Answer: C.

In fact we arrived to the answer C, without using the info about the intersection point of the lines. So this info is not needed to get C.

For more on coordinate geometry check the link in my signature.



Bunuel i just dont understand "But from this statement we can deduce the following: x-intersect is value of for " this part rest of them are ok. Can you tell me how did you decide y=0?
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Re: In the xy-coordinate plane, line l and line k intersect at the point  [#permalink]

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New post 07 Sep 2010, 08:52
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fatihaysu wrote:
Bunuel i just dont understand "But from this statement we can deduce the following: x-intersect is value of for " this part rest of them are ok. Can you tell me how did you decide y=0?


X-intercept is the point where a line (a graph) crosses the x-axis. So it's the point on x-axis, any point on x-axis has y-coordinate equal to zero, which means that X-intercept is the point \((x,0)\) - the value of \(x\) when \(y=0\): \(y=mx+b\) --> \(0=mx+b\) --> \(x=-\frac{b}{m}\). So X-intercept of a line \(y=mx+b\) is \(x=-\frac{b}{m}\);

Y-intercept is the point where a line (a graph) crosses the y-axis. So it's the point on y-axis, any point on y-axis has x-coordinate equal to zero, which means that Y-intercept is the point \((0,y)\) - the value of \(y\) when \(x=0\): \(y=mx+b\) --> \(y=m*0+b\) --> \(y=b\). So Y-intercept of a line \(y=mx+b\) is \(y=b\).

Check Coordinate Geometry chapter for more (link in my signature).

Hope it's clear.
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Re: GMAT PREP (DS)  [#permalink]

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New post 25 Sep 2010, 12:27
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The equation of a line can be written as :

\(\frac{x}{a} + \frac{y}{b} = 1\)

Here a is the X-intercept and b is the Y-intercept

The slope of this line is -b/a

Let the lines be :

\(\frac{x}{a_1} + \frac{y}{b_1} = 1\)
\(\frac{x}{a_2} + \frac{y}{b_2} = 1\)

We know both lines pass through (4,3)
So we know that
\(\frac{4}{a_1} + \frac{3}{b_1} = \frac{4}{a_2} + \frac{3}{b_2} = 1\)
The information about (4,3) doesn't tell us much about the signs of a1,a2,b1,b2

Question is if Slope1*Slope2 =(b1 b2) / (a1 a2) < 0

(1) a1 * a2 > 0
Depends on the sign of b1*b2

(2) b1 * b2 < 0
Depends on the sign of a1*a2

(1) + (2) : Clearly sufficient to say yes. The slope product is always < 0.
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Re: GMAT PREP (DS)  [#permalink]

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New post 27 Sep 2010, 08:39
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under timed schedule, i considered using a quick graphical sketch and that helped.
Let M1 and M2 represent the gradients of lines K and L

(1)
from sketch (i), M1*M2 < 0
from sketch (ii), M1*M2 > 0
INSUFFICIENT

(2)
from sketch (i), M1*M2 < 0
from sketch (iii), M1*M2 > 0
INSUFFICIENT

Combining (1) and (2):
ONLY sketch i) satisfies --> SUFFICIENCY
option C is therefore correct.
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Re: GMAT PREP (DS)  [#permalink]

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New post 27 Apr 2011, 11:54
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1
In the xy-coordinate plant, line t and k intersect at (4,3). Is the product of their slopes negative?
1. Product of the x intercepts of the line t & k is positive
2. Product of the y intercepts of the line t & k is negative

Sol:
To draw a line with +ve slope through a point(x1, y1) on a 2D plane; draw two lines passing through the point perpendicular to each other with one line parallel to x-axis and the other parallel to y-axis(depicted by red lines in the images). These red-lines make 4-quadrants with origin(x1,y1). Any line that passes through this origin and lies on 1st and 3rd quadrant will have +ve slope. Any line passing through the point and lying on the 2nd and 4th quadrants will have negative slope.

Attached images show two possible scenarios for St1 and St2 making the statements insufficient individually. Last image shows the only possible scenario for the lines, making both the statements sufficient together.

St1:
Insufficient.
Attachment:
product_slopes_of_two_lines_3_n_4_St1.PNG
product_slopes_of_two_lines_3_n_4_St1.PNG [ 14.96 KiB | Viewed 13922 times ]



St2:
Insufficient.
Attachment:
product_slopes_of_two_lines_3_n_4_St2.PNG
product_slopes_of_two_lines_3_n_4_St2.PNG [ 15.7 KiB | Viewed 13914 times ]


Combined:
Sufficient.
Attachment:
product_slopes_of_two_lines_3_n_4_Combined.PNG
product_slopes_of_two_lines_3_n_4_Combined.PNG [ 12.94 KiB | Viewed 13887 times ]


Ans: "C"
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Re: GMAT PREP (DS)  [#permalink]

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New post 27 Apr 2011, 20:13
(1)

So they both have +ve x-intercept or -ve x=intercept

This is not sufficient, as can be seen from diagram, both can have +ve slope or 1 can have -ve and 1 +ve slope

(2)

So they both intersect y-axis on different sides of X-Axis

This is also not sufficient, as can be seen from diagram, both can have +ve slope or 1 can have -ve and 1 +ve slope

(1) and (2)

They are sufficient, as only the option with two lines on right of y-axis is possible.

Hence products of slopes is negative.

Answer - C
Attachments

Lines.JPG
Lines.JPG [ 16.69 KiB | Viewed 13090 times ]


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Re: GMAT PREP (DS)  [#permalink]

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New post 26 May 2012, 01:11
Bunuel wrote:
LM wrote:
Please explain...


In the xy coordinate plane, line L and line K intersect at the point (4,3). Is the product of their slopes negative?

We have two lines: \(y_l=m_1x+b_1\) and \(y_k=m_2x+b_2\). The question: is \(m_1*m_2<0\)?

Lines intersect at the point (4,3) --> \(3=4m_1+b_1\) and \(3=4m_2+b_2\)

(1) The product of the x-intersects of lines L and K is positive. Now, one of the lines can intersect x-axis at 0<x<4 (positive slope) and another also at 0<x<4 (positive slope), so product of slopes also will be positive BUT it's also possible one line to intersect x-axis at 0<x<4 (positive slope) and another at x>4 (negative slope) and in this case product of slopes will be negative. Two different answers, hence not sufficient.

But from this statement we can deduce the following: x-intersect is value of \(x\) for \(y=0\) and equals to \(x=-\frac{b}{m}\) --> so \((-\frac{b_1}{m_1})*(-\frac{b_2}{m_2})>0\) --> \(\frac{b_1b_2}{m_1m_2}>0\).

(2) The product of the y-intersects of lines L and K is negative. Now, one of the lines can intersect y-axis at 0<y<3 (positive slope) and another at y<0 (positive slope), so product of slopes will also be positive BUT it's also possible one line to intersect y-axis at y<0 (positive slope) and another at y>3 (negative slope) and in this case product of slopes will be negative. Two different answers, hence not sufficient.

But from this statement we can deduce the following: y-intercept is value of \(y\) for \(x=0\) and equals to \(x=b\) --> \(b_1*b_2<0\).

(1)+(2) \(\frac{b_1b_2}{m_1m_2}>0\) and \(b_1*b_2<0\). As numerator in \(\frac{b_1b_2}{m_1m_2}>0\) is negative, then denominator \(m_1m_2\) must also be negative. So \(m_1m_2<0\). Sufficient.

Answer: C.

In fact we arrived to the answer C, without using the info about the intersection point of the lines. So this info is not needed to get C.

For more on coordinate geometry check the link in my signature.

P.S. This question can be easily solved by drawing the lines without any calculations.


great explanation....

i just plotted the lines on a coordinate figure....

st 1: product of x-intercepts are positive, so from point (4,3) in first quadrant we could have two possibilities:
a) 1st line - negative slope and a positive x-intercept and 2nd line - positive slope and positive x-intercept (y-intercepts positive - line 1 and negative = line 2)
b) 1st line - positive slope and a positive x-intercept and 2nd line - positive slope and positive x-intercept (y-intercepts negative)

hence 2 cases are possible... not sufficient

st 2: approach the statement with the same philosophy as in st 1 and arrived at two cases,
a) 1st line - negative slope and a positive y-intercept and 2nd line - positive slope and negative y-intercept (x-intercepts positive)
b) 1st line - positive slope and a negative y-intercept and 2nd line - positive slope and positive y-intercept (x-intercepts positive - line 1 and negative - line 2)

hence 2 cases are possible... not sufficient

from both the statements, we can arrive at a unique case case a) from st 1 and case a) from st 2

answer therefore is C

the cases look very much conceivable on a graph plot.
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In the xy coordinate plane, line L and line K intersect at the p  [#permalink]

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New post 10 May 2013, 12:58
M8 wrote:
In the xy-coordinate plane, line l and line k intersect at the point (4,3). Is the product of their slope negative?

1) The product of the x-intercepts of lines l and k is positive.
2) The product of the y-intercepts of lines l and k is negative.

I cracked this question right. :wink:
But these types of questions are real horror for me.

Guys please elaborate your solution for me. Thank you.

I'll post the OA later.


Let X1 and X2 are the X intercepts for the lines l and K.

Calculate slope for L [(4,3) and (X1, 0)] and m [ (4,3) and (X2, 0)] such that X1*X2=+VE

Similarly take Y1 and Y2. and calculate slope for L [(4,3) and (0, Y1)] and m [ (4,3) and (0, Y2)] such that Y1*Y2=-VE

Now combine these two statement and calculate slope for L [(0, Y1) and (X1, 0)] and m [ (0, Y2) and (X2, 0)]

Slope of L*M = Y1/(-X1)*Y2/(-X2) = Y1*Y2/ X1*X2.

Given that Y1*Y2= -VE and X1*X2=+VE.

Hence Slope of L*M = -VE

Answer C.
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Re: In the xy coordinate plane, line L and line K intersect at  [#permalink]

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New post 27 Jun 2013, 22:28
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Re: In the xy coordinate plane, line L and line K intersect at  [#permalink]

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New post 25 Nov 2014, 08:01
shrouded1 wrote:
The equation of a line can be written as :

\(\frac{x}{a} + \frac{y}{b} = 1\)

Here a is the X-intercept and b is the Y-intercept

The slope of this line is -b/a

Let the lines be :

\(\frac{x}{a_1} + \frac{y}{b_1} = 1\)
\(\frac{x}{a_2} + \frac{y}{b_2} = 1\)

We know both lines pass through (4,3)
So we know that
\(\frac{4}{a_1} + \frac{3}{b_1} = \frac{4}{a_2} + \frac{3}{b_2} = 1\)
The information about (4,3) doesn't tell us much about the signs of a1,a2,b1,b2

Question is if Slope1*Slope2 =(b1 b2) / (a1 a2) < 0

(1) a1 * a2 > 0
Depends on the sign of b1*b2

(2) b1 * b2 < 0
Depends on the sign of a1*a2

(1) + (2) : Clearly sufficient to say yes. The slope product is always < 0.


Did this way using the intercept form of a line. Have made an observation that using intercept equations in gmat questions significantly reduces the time to solve a question.
Thanks.

A thumb rule I use in co-ordinate geometry for lines:
1. Use intercept form x/a + y/b = 1 - If intercepts are given, for area questions usually, for ladder against wall type questions, triangle questions
2. Use slope form y=mx+c - when slope is implicit or explicitly stated in the question
3. Use one point slope form y-y1=m(x-x1) - when we know slope and line passes through a point
4. Use 2 point form y-y1/y2-y1 = x-x1/x2-x1 - when line passes through two points
5. When all else fails we resort to pure algebra of a linear equation ax+by+c=0
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Re: In the xy-coordinate plane, line l and line k intersect at the point  [#permalink]

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New post 02 Jun 2015, 03:13
M8 wrote:
In the xy-coordinate plane, line l and line k intersect at the point (4,3). Is the product of their slope negative?

1) The product of the x-intercepts of lines l and k is positive.
2) The product of the y-intercepts of lines l and k is negative.

.


Question: Is the product of slopes of line L and K negative?

Requirement: For the product of the slopes to be negative one of the lines must have positive slope (Sloping upward from left to right) and one of the lines must have Negative slope (Sloping Downward from left to right)

Statement 1: The product of the x-intercepts of lines l and k is positive.

With positive X intercept of both the lines, both the lines may have positive slope
OR
With positive X intercept of both the lines, One of the lines may have positive slope and other line may have negative slope therefore NOT SUFFICIENT


Statement 2: The product of the y-intercepts of lines l and k is negative.

With positive Y intercept of one of the lines and Negative Y intercept of other line, both the lines may have positive slope
OR
With positive Y intercept of one of the lines and Negative Y intercept of other line, One of the lines may have positive slope and other line may have negative slope therefore NOT SUFFICIENT

Combining the two Statements:

Case 1: Both Lines have positive X intercepts (For product of therefore X intecepts to be (+ve) )
One of the lines will have positive Y intercept and one Line will have Negative Y intercept

The line with (+ve) X-intecept and (+ve)Y intercept will bound to have (-ve)Slope
and
The line with (+ve) X-intecept and (-ve)Y intercept will bound to have (+ve)Slope
Hence Product of slopes will be Negative


Case 2: Both Lines have Negative X intercepts (For product of therefore X intecepts to be (+ve) )
One of the lines will have positive Y intercept and one Line will have Negative Y intercept

The line with (-ve) X-intecept and (+ve) Y intercept will bound to have (+ve)Slope
and
The line with (-ve) X-intecept and (-ve) Y intercept will bound to have (-ve)Slope
Hence Product of slopes will be Negative

Consistent answer
therefore, SUFFICIENT

Answer: Option
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Re: In the xy-coordinate plane, line l and line k intersect at the point  [#permalink]

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New post 09 Jun 2016, 23:26
If we modify the original condition and the question, the equation for a line that crosses (2,1) and (4,3) is y=x-1. Then, if B(m,n), n=m-1. There are, 2 variables (m and n) and 1 equation (n=m-1). In order to match the number of variables to the number of equations, we need 1 equation. Since the condition 1) and the condition 2) each has 1 equation, there is high chance that D is the correct answer.
Since 1)=2) (m,n)=(3,2), the answer is yes and the conditions are sufficient. The correct answer is D.
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Re: In the xy coordinate plane, line L and line K intersect at  [#permalink]

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New post 13 Nov 2016, 17:29
Can the question be solved the following way? Or, am I just lucky with the solution particular to this question?

Slope of a line = \(\frac{rise}{run}\) = \(\frac{y-intercept}{x-intercept}\)

With this in mind,

(1) The product of the x-intercepts of lines \(l\) and \(k\) is positive

Translation:

\(\frac{y}{x}\) = \(+\) or \(\frac{-y}{x}\) = \(-\)

Insufficient

(2) The product of the y-intercepts of lines \(l\) and \(k\) is negative

Translation:

\(\frac{-y}{-x}\) = \(+\) or \(\frac{-y}{x}\) = \(-\)

Insufficient

(1) & (2):

\(\frac{-y}{x}\) = \(-\) is the only answer in common and therefore, sufficient.

Please advise,

Thank you!
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Re: In the xy-coordinate plane, line l and line k intersect at the point  [#permalink]

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New post 28 Aug 2017, 08:24
2
M8 wrote:
In the xy-coordinate plane, line l and line k intersect at the point (4,3). Is the product of their slope negative?

1) The product of the x-intercepts of lines l and k is positive.
2) The product of the y-intercepts of lines l and k is negative.

I cracked this question right. :wink:
But these types of questions are real horror for me.

Guys please elaborate your solution for me. Thank you.

I'll post the OA later.


Always draw the figure to solve such questions

Answer option C
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Re: In the xy-coordinate plane, line L and link K intersect at  [#permalink]

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New post 26 Sep 2019, 02:54
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Re: In the xy-coordinate plane, line L and link K intersect at   [#permalink] 26 Sep 2019, 02:54
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